Ta có: A = 1/2+1/3+1/4+...+1/62+1/63+1/64

A = 1+(1/2+1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+...+1/16)+...+(1/17+1/18+....+1/32)+(1/33+1/34+...+1/64)

Ta có: 1/2+1/3+1/4>1/2+1/4+1/4=1

1/5+1/6+1/7+1/8>1/8+1/8+1/8+1/8=1/8.4=1/2

1/9 +1/10+...+1/16>1/16+1/16+...1/16=1/16.8=1/2

1/33+1/34+...+1/64>1/64+1/64+...+1/64=1/64.32=1/2

Vậy A > 4

Xin ai giải hộ cái

Ta có A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/64

          A = 1 + (1/2 + 1/3 + 1/4) + (1/5 + 1/6 + ... + 1/8) + (1/9 + 1/10 + 1/11 + ... + 1/16) + (1/17 + 1/18 + 1/19 + ... + 1/32) + (1/33 + 1/34 + 1/35 + ... + 1/64)

=> A > 1 +  (1/2 + 1/4.2) + 1/8.4 + 1/16.8 + 1/32.16 + 1/64.32

     A > 1 + 1 + 1/2 + 1/2 + 1/2 + 1/2

    A > 4 (DPCM).

A=1+(1/2 + 1/3 + 1/4)+(1/5 + 1/6 + 1/7 + 1/8)+(1/9+...+1/16)+(1/17+...+1/32)+(1/33+...+1/64)

A>1+(1/2 + 1/4 + 1/4)+(1/8+ 1/8+ 1/8+ 1/8)+(1/16+1/16+...+1/16)+(1/64+...+1/64)

A>1 + 1 + 1/2 + 1/2 + 1/2+ 1/2

A>4

\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)

\(=1+\frac{1}{2}+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32\)

\(=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

\(=1+\frac{1}{2}.6\)

\(=1+3\)

\(=4\)

~~ Bố thí cái li.ke ~~

Ai làm hộ mình với